Reflection, Transmission, and Dispersion

How boundary conditions determine reflection phase and amplitude ratios, and how dispersion differs from an ordinary change in wavelength.

Reflection and transmission concern what happens at an interface. Dispersion concerns how propagation speed depends on frequency. They are different questions, although both can occur in the same experiment.

1. Start with a travelling wave

Write a sinusoidal wave travelling in the positive $𝑥$ direction as

𝑢 (𝑥, 𝑡) = 𝐴 cos (𝑘 𝑥 - 𝜔 𝑡 + 𝜑) .

Here $𝑢$ is the quantity that oscillates, such as string displacement or an electric-field component. The amplitude is $𝐴$ , the angular frequency is $𝜔 = 2 𝜋 𝑓$ , the wavenumber is $𝑘 = 2 \frac{𝜋}{𝜆}$ , and $𝜑$ is a constant phase offset. Frequency $𝑓$ counts oscillations per second; wavelength $𝜆$ is the distance between successive crests.

To follow one crest, keep its phase constant:

𝑘 𝑥 - 𝜔 𝑡 + 𝜑 = constant \Rightarrow 𝑣_{𝑝} = \frac{𝑑 𝑥}{𝑑 𝑡} = \frac{𝜔}{𝑘} = 𝑓 𝜆 .

The speed $𝑣_{𝑝}$ is called the phase velocity: the speed of a fixed phase of a sinusoidal wave. This does not yet tell us whether the medium is dispersive.

2. Boundary conditions determine reflection and transmission

Suppose a wave in medium 1 meets medium 2 at $𝑥 = 0$ . There is an incident wave approaching the interface, a reflected wave travelling back into medium 1, and a transmitted wave travelling into medium 2.

The total wave must satisfy the physical constraints at that interface. For joined strings, the displacement is continuous and the transverse forces must balance. For light at an ordinary interface with no imposed surface current, the tangential electric and magnetic fields are continuous.

Solving these boundary conditions gives the amplitude coefficients

𝑟 = \frac{𝐴_{𝑟}}{𝐴_{𝑖}}, 𝜏 = \frac{𝐴_{𝑡}}{𝐴_{𝑖}} .

The subscripts $𝑖$ , $𝑟$ , and $𝑡$ mean incident, reflected, and transmitted. The transmission coefficient is often written $𝑡$ ; here we use $𝜏$ to avoid confusing it with time. These are signed amplitudes, or complex amplitudes when a general phase difference is needed, not just positive wave heights.

2.1. The sign of the reflection coefficient gives the inversion

Use the same time dependence for all waves. For real coefficients, a convenient choice is

\begin{aligned} 𝑢_{𝑖} & = 𝐴_{𝑖} cos (𝑘_{1} 𝑥 - 𝜔 𝑡), \\ 𝑢_{𝑟} & = 𝑟 𝐴_{𝑖} cos (- 𝑘_{1} 𝑥 - 𝜔 𝑡), \\ 𝑢_{𝑡} & = 𝜏 𝐴_{𝑖} cos (𝑘_{2} 𝑥 - 𝜔 𝑡) . \end{aligned}

At the boundary, $𝑢_{𝑖} (0, 𝑡) = 𝐴_{𝑖} cos (𝜔 𝑡)$ and $𝑢_{𝑟} (0, 𝑡) = 𝑟 𝐴_{𝑖} cos (𝜔 𝑡)$ . Therefore:

$𝑟 > 0$ : the reflected oscillation is in phase with the incident oscillation at the boundary. There is no inversion.
$𝑟 < 0$ : the reflected oscillation has the opposite sign. Since $- cos (𝜃) = cos (𝜃 + 𝜋)$ , this is a $𝜋$ -radian, or $180 °$ , phase shift.
$𝑟 = 0$ : there is no reflected wave, so its phase is undefined.

The reflected amplitude magnitude is $| 𝑟 | 𝐴_{𝑖}$ in either nonzero case. For complex coefficients, write $𝑟 = | 𝑟 | 𝑒^{𝑖 𝜑_{𝑟}}$ ; the reflection phase is $𝜑_{𝑟} = arg (𝑟)$ , not necessarily just zero or $𝜋$ .

Compare phases at the interface using a consistent wave convention. An arbitrary phase parameter in a sine formula is not automatically the physical reflection phase. For example, at a fixed end placed at $𝑥 = 0$ ,

𝐴 sin (𝑘 𝑥 - 𝜔 𝑡) + 𝐴 sin (𝑘 𝑥 + 𝜔 𝑡) = 0 at 𝑥 = 0 .

No extra $𝜋$ appears in that particular formula, but the reflected oscillation at the boundary is still the negative of the incident one. The opposite signs of the time arguments already supply the inversion.

2.2. Fixed and free ends of a string

At a fixed end, the total displacement must be zero:

𝑢_{𝑖} (0, 𝑡) + 𝑢_{𝑟} (0, 𝑡) = 0 \Rightarrow 𝑟 = - 1 .

The reflected displacement has the same amplitude magnitude and is inverted.

At an ideal free end, the transverse force vanishes. For a string under tension, this means its slope at the end is zero:

\frac{𝜕 𝑢}{𝜕 𝑥} (0, 𝑡) = 0 \Rightarrow 𝑟 = + 1 .

The reflected displacement has the same amplitude magnitude and is not inverted.

A perfect electrical conductor imposes zero total tangential electric field at its surface. At normal incidence this similarly gives $𝑟_{𝐸} = - 1$ : the reflected electric field is inverted. Perfect reflection alone does not specify a phase; the boundary condition and the oscillating quantity do.

3. Impedance mismatch determines the amplitude ratios

Impedance describes the relation between the two coupled quantities that carry a travelling wave, such as transverse force and velocity on a string, or electric and magnetic fields for light. Matching those quantities at the boundary determines how much is reflected and transmitted.

3.1. Joined strings: displacement coefficients

For ideal strings under the same tension $𝐹$ , let $𝜇_{𝑗}$ be the mass per unit length of string $𝑗$ . Its wave speed and characteristic impedance are

𝑣_{𝑗} = \sqrt{\frac{𝐹}{𝜇_{𝑗}}}, 𝑍_{𝑗} = 𝜇_{𝑗} 𝑣_{𝑗} = \sqrt{𝐹 𝜇_{𝑗}} .

Continuity of displacement and balance of transverse force give

1 + 𝑟 = 𝜏, 𝑍_{1} (1 - 𝑟) = 𝑍_{2} 𝜏 .

Solving gives

𝑟 = \frac{𝑍_{1} - 𝑍_{2}}{𝑍_{1} + 𝑍_{2}}, 𝜏 = \frac{2 𝑍_{1}}{𝑍_{1} + 𝑍_{2}} .

Thus displacement reflection toward a higher string impedance is inverted; reflection toward a lower string impedance is not. Equal impedances give $𝑟 = 0$ and $𝜏 = 1$ .

These are coefficients for displacement. Do not copy their signs blindly into a pressure-wave problem. For acoustic pressure, using acoustic impedance $𝑍 = 𝜌 𝑐_{𝑠}$ with density $𝜌$ and sound speed $𝑐_{𝑠}$ , the normal-incidence reflection coefficient is instead $\frac{𝑍_{2} - 𝑍_{1}}{𝑍_{2} + 𝑍_{1}}$ . Pressure and particle velocity have different reflection signs.

3.2. Light: normal-incidence Fresnel coefficients

For ordinary, lossless, nonmagnetic media, let $𝑛_{1}$ and $𝑛_{2}$ be the real, positive refractive indices. At normal incidence, measure all electric-field amplitudes along the same tangential axis. The boundary conditions give

1 + 𝑟 = 𝜏, 𝑛_{1} (1 - 𝑟) = 𝑛_{2} 𝜏 .

The resulting Fresnel coefficients are

𝑟 = \frac{𝑛_{1} - 𝑛_{2}}{𝑛_{1} + 𝑛_{2}}, 𝜏 = \frac{2 𝑛_{1}}{𝑛_{1} + 𝑛_{2}} .

Consequently:

Lower to higher refractive index: $𝑟 < 0$ , so the reflected electric field is inverted.
Higher to lower refractive index: $𝑟 > 0$ , so it is not inverted.
Equal indices: $𝑟 = 0$ , so there is no reflected wave under these assumptions.

The electromagnetic wave impedance in these media is $𝜂_{𝑗} = \frac{𝜂_{0}}{𝑛_{𝑗}}$ , where $𝜂_{0}$ is the vacuum impedance. Equivalently, $𝑟 = \frac{𝜂_{2} - 𝜂_{1}}{𝜂_{2} + 𝜂_{1}}$ . Notice that increasing refractive index means decreasing electromagnetic impedance. This is why a universal mnemonic such as “higher impedance always inverts the wave” is unsafe without identifying the wave quantity.

These formulas assume normal incidence. At oblique incidence the Fresnel coefficients also depend on angle and polarization. Absorbing media and total internal reflection can give complex reflection coefficients and phase shifts other than zero or $𝜋$ .

3.3. Amplitude ratios are not power fractions

Energy conservation applies to energy flux, not to a sum of amplitudes. For the lossless string problem above, the reflected and transmitted fractions of incident power are

𝑅 = {| 𝑟 |}^{2}, 𝑇 = (\frac{𝑍_{2}}{𝑍_{1}}) {| 𝜏 |}^{2} .

For the normal-incidence optical problem above, they are

𝑅 = {| 𝑟 |}^{2}, 𝑇 = (\frac{𝑛_{2}}{𝑛_{1}}) {| 𝜏 |}^{2} .

In both cases, $𝑅 + 𝑇 = 1$ . The extra transmission factor accounts for the fact that the same amplitude can carry different power in different media.

For example, light passing from air with $𝑛_{1} = 1$ into glass with $𝑛_{2} = 1.5$ has

𝑟 = - 0.2, 𝜏 = 0.8, 𝑅 = 0.04, 𝑇 = 0.96 .

The reflected electric field is inverted and has one fifth of the incident amplitude, but it carries only $4 %$ of the incident power. The transmitted field has $80 %$ of the incident amplitude and carries $96 %$ of the power. In the reverse direction $𝜏 = 1.2$ , yet $𝑇 = 0.96$ still: an amplitude ratio larger than one does not by itself imply an energy gain.

4. Dispersion: does phase speed depend on frequency?

Wavelength depends on frequency even in a nondispersive medium. The decisive question is whether phase speed also depends on frequency.

4.1. No dispersion

If every frequency has the same phase speed $𝑣$ , then

𝜔 = 𝑣 𝑘, 𝜆 = \frac{2 𝜋 𝑣}{𝜔} .

Higher frequency means shorter wavelength, but all those sinusoidal components travel at the same speed. This inverse dependence of wavelength on frequency is not dispersion. Ideal light propagation in vacuum is an example: $𝑣 = 𝑐$ and $𝜆 = 2 𝜋 \frac{𝑐}{𝜔}$ .

4.2. Dispersion

A dispersion relation is the equation relating frequency and wavenumber, usually written $𝜔 = 𝜔 (𝑘)$ . A dispersion relation exists even for a nondispersive wave; its name does not imply that the wave is dispersive.

When phase speed varies with frequency,

𝑣_{𝑝} (𝜔) = \frac{𝜔}{𝑘 (𝜔)}, 𝜆 (𝜔) = \frac{2 𝜋 𝑣_{𝑝} (𝜔)}{𝜔} .

For light in a transparent medium,

𝑣_{𝑝} (𝜔) = \frac{𝑐}{𝑛 (𝜔)}, 𝜆 (𝜔) = \frac{2 𝜋 𝑐}{𝑛 (𝜔) 𝜔} .

A frequency-dependent refractive index therefore produces dispersion. Different frequency components accumulate different propagation phases and a wave packet can spread or otherwise change shape.

For a narrowband packet, the envelope travels approximately at the group velocity

𝑣_{𝑔} = \frac{𝑑 𝜔}{𝑑 𝑘} .

In a nondispersive medium with $𝜔 = 𝑣 𝑘$ , both $𝑣_{𝑔}$ and $𝑣_{𝑝}$ equal $𝑣$ . In a dispersive medium they generally differ. Group velocity describes the packet envelope under the narrowband approximation; it is not automatically the speed of information in every propagation regime.

5. Changing wavelength at a boundary is not itself dispersion

At a stationary, linear, time-independent interface, the incident, reflected, and transmitted waves have the same frequency. Boundary conditions must hold at every instant; unrelated sinusoidal frequencies cannot maintain those conditions for all time. Moving or time-modulated interfaces are outside this assumption.

The transmitted speed can differ from the incident speed, so its wavelength changes according to

𝜆_{1} = \frac{𝑣_{𝑝, 1}}{𝑓}, 𝜆_{2} = \frac{𝑣_{𝑝, 2}}{𝑓} .

For example, monochromatic light of vacuum wavelength $600 nm$ entering glass with $𝑛 = 1.5$ has wavelength $400 nm$ inside the glass. Its frequency is unchanged. That one wavelength change does not establish dispersion: to check for dispersion, compare the medium’s phase speed at different frequencies.

An interface can also reflect different frequencies by different amounts if its coefficients depend on frequency. That is a frequency-dependent boundary response, distinct from propagation dispersion within the medium.

Quick decision guide

Will the reflected wave invert? Identify the oscillating quantity and boundary conditions, then check the sign of $𝑟$ (or $arg (𝑟)$ if complex).
How large are the reflected and transmitted waves? Calculate $| 𝑟 |$ and $| 𝜏 |$ from the appropriate impedance or Fresnel formulas.
How much power goes each way? Use $𝑅$ and $𝑇$ , including the appropriate impedance or refractive-index factor.
Is there dispersion? Check whether $𝑣_{𝑝} = \frac{𝜔}{𝑘}$ depends on frequency, not merely whether $𝜆$ changes with frequency.
Did the wavelength change on entering another medium? Use $𝑣_{𝑝} = 𝑓 𝜆$ with the same frequency on both sides of a stationary, time-independent interface. That change alone is not evidence of dispersion.